I am a high school student doing a science project in superconductivity. My knowledge in physics is hence not pretty advanced but I am doing my best to understand the subject. What I find really difficult to understand is "the current" inside/ outside the superconductor. I've read that a current on the superconductor is induced in order to repel the magnetic field from the magnet because of a change in the magnetic flux (others says that the current is induced to prevent the magnetic flux from changing) . But this current (eddy current?) is made of electrons right?. I've read that there are essentially two sorts of electrons in a superconductors, quote from a book

"a superconductor below its transition temperature appears to be permeated by two electron fluids, one of normal electrons and one of super electrons."

What does this mean? are these super electrons "cooper pairs"? This induced current is made of cooper pairs? I would really appreciate help in this. There is no clear information and I really don't want to give up my project. :)


Superconductivity is one of those topics where it's really easy to give an intuitive & wrong explanation. One can go with a simplified picture at first, but when you start understanding it, you are going to come up with more questions, the answers to which will start undoing the (incorrect) premises to the initial, intuitive, answer.

So, I will give a very long answer which hopefully covers all of our bases.

First of all, what I will explain as follows applies to conventional superconductors, that is the ones that are currently understood. Unconventional superconductors are more prominent in the news nowadays because they are the ones with high temperature transition points, but there is not yet an agreed theory describing them.

Normal conductors

First off, particles bound in a potential (i.e. not free running in free space) will have discrete energy levels. Electrons in a conductor are bound, that is you would have to provide energy to get them out of the conductor (this is called the work function and it's important in the photoelectric effect). The force providing the binding force is whatever is holding the solid together, i.e. making it into a crystalline solid. As a result, electrons will sit in a discrete ladder of energy levels.

Electrons are fermions, which means they obey the Pauli exclusion principle, which essentially forbids two identical fermions from existing. "Identcial" would mean electrons at the same place, with the same velocity, with the same spin and spin orientation. So, at absolute 0 temperature (when you have the "real" ground state), while bosons would bunch up in the lowest energy level (sometimes triggering Bose-Einstein condensation), fermions need to stack up -- the two colours here refer to the spin orientation of the electron, which can either be up or down (From here):

enter image description here

The highest occupied energy level is known as the Fermi level (technically it's the difference between the highest occupied and the lowest unoccupied). The key thing about a metal is that this level occurs within a band -- meaning that the states above that level are still accessible. In an insulator, there would be a gap for example. But in a metal, as soon temperature is non-zero, electrons in this top level can acquire thermal energy $k_B T$ and jump up to some other levels.

This is usually depicted like this (image from here):

enter image description here

where the valence band is essentially all the levels below the Fermi level, and the conduction band are the ones above.

The key thing to understand is that, when a band is filled, it means that electrons are occupying all possible available states. If there is a state with velocity $+v$, there is also one with velocity $-v$ -- and if all are filled, then the net speed of the electrons has to be zero. That is, no electrical current can flow!

The importance of having thermally excited electrons into the conduction band, is that there isn't a counterpart at the opposite velocity -- so now you can actually have a net charge with a net velocity, which can carry electricity.

Hence even a normal conductor would have zero resistance, but only at 0 temperature (unless there are impurities but that's not temperature dependent).


Now, superconductivity arises because of an interaction between electrons and the crystal lattice of the solid ("electron-phonon interaction") stylised here:

enter image description here

This interaction is very weak, so it can only form at very low temperatures where the binding energy of this state is less than $k_B T$ so it cannot be destroyed by thermal fluctuations.

Indeed, this binding energy $\Delta$ scales as temperature $T$:

$$\Delta(T) \propto \left ( 1-\frac{T}{T_c} \right )^{\text{some power}}, $$

and is used as the order parameter for superconductivity. The order parameter is something that quantifies a new phase, meaning that $\Delta = 0$ for $T>T_c$ and $\Delta > 0$ for $T < T_c$, where $T_c$ is the critical temperature.

This bound state is known as a Cooper pair.

Whether or not it can be interpreted as two electrons is best explained by John Bardeen himself (the 'B' in the BCS theory of superconductivity):

"The idea of paired electrons, though not fully accurate, captures the sense of it."

This is maybe too advanced for you, but the 'electron' as we think about it really is the "free" electron, on its own. Its character changes when highly interacting & correlated with another one. This is a quantum thing. But yeah, the electrons in the Cooper pair are what your book refers to as "super electrons".

Being a bound state, the Cooper pair has a lower energy than two free electrons. That is, you'd have to provide energy to the Cooper pair to break it and recover the two free electrons.

Because of this, the energy of the "electron cloud" (meaning all the electrons in the metal) goes down as more Cooper pairs are formed (that is as we reduce the temperature further). A gap opens in the band structure, as shown here:

enter image description here

In the superconducting picture above, you now have a band gap that divides the electrons into 2 bands now. The gap is larger than the temperature fluctuations, so it's very difficult if not impossible for electrons to jump to the band above. The left band includes both free electrons and Cooper pairs. As we said for normal conductors, however, electrons in a full band cannot have a net velocity, since for any $-v$ occupied state, there will be a $+v$ occupied state.

So these electrons cannot move around and cannot conduct electricity and form your surface currents! But, in this band, there are also "super electrons" that have formed Cooper pairs. Cooper pairs are not strictly speaking fermions anymore, so they don't have to obey the Pauli exclusion principle: more than one Cooper pair can have the same velocity $v$, which adds up and gives you a macroscopic current. Cooper pairs conduct electricity in the superconducting state, and will be the ones generating the surface currents you mention.

Now to address your quotation, yes as already mentioned there are indeed two "kinds" of electrons in this band -- the electrons that are in Cooper pairs ("super electrons") and the ones that are not ("normal"). This allows the introduction of a two fluid model, where the electron density is divided into $n_s$ and $n_n$ as shown here:

enter image description here

This picture is however misleading as it suggests that there's $n_0$ normal electrons at $T_c$ and they are all converted into Cooper pairs ("super electrons"). To correct this, in the text they say:

Here $n_0$ is the superconducting electron density at $T = 0$ K.

which is quite sneaky, but basically means they are going "the other way around": whatever Cooper pairs there are at $T=0$, they call that $n_0$, and they say that these would be normal electrons at $T=T_c$.

Indeed, it is tempting to think that, at $T=0$, all electrons become Cooper pairs. But this is not true. You have to satisfy very precise relations for electrons to form a Cooper pair, which are usually only satisfied near the Fermi level so for energies $\epsilon_F \pm \Delta$.

However, because the band is filled, the normal electrons do not "move" -- only Cooper pairs conduct electricity.

Perfect diamagnetism

This is now a very subtle point.

Suffice to say, nature abhors changes in magnetic field (technically magnetic flux). So if you have a current carrying wire, and ramp up an external magnetic field, the charges in the wire will try to rearrange themselves so that they can generate a field that is equal and opposite, in order for the total field to be zero (or at least equal to the one purely generated by the wire, which was there even before we ramped up the external one).

But notice that this arises only because there was a change in the external field. If I had had a magnetic field on from the beginning, and then started a current in the wire, this would not happen.

This is the real, experimental tell-tale that it is a superconductor and not just an "ideal" (zero resistance) conductor: whatever magnetic field inside of it, it gets expelled, i.e. the Meissner effect. Even if it was present prior to current flowing. Note that an ideal conductor would not expel it, it will allow it if it was present before current was flowing. But superconductors expel it no matter what.

And this is why a superconductor is not just an ideal conductor.

The reason for this expulsion is due to the superconductor going through a phase transition in which the microscopic description of the material changes. It is well beyond high school level, but the electromagnetic field propagates in a different way when the material becomes superconducting, and it follows the London equations, which show an exponential damping of the field.

Of course, it is surface currents that are making sure their field is equal and opposite the external field, so as to ensure the sum is zero inside the superconductor. And as said before, the only charges that are able to move are Cooper pairs. So yes, Cooper pairs are forming surface currents to repel external fields.

  • $\begingroup$ Thank you so much for taking your time to answer!! This really helped a lot. I have another question if it isn't too much to ask. I understand that when a magnetic field is first applied and then the supercondutor reaches its critical T. The magnet will be repelled due to the Meissner effect. But if the superconductor first is cooled down to its critical T and then the magnetic field is applied. In that case, Is the magnet repelled due to Lenz's law? $\endgroup$ Oct 11 '21 at 19:46
  • 1
    $\begingroup$ That's a good point and honestly I don't know. Cooper pairs are free charges so they will "respond" to Lenz' law after all - so maybe that's a contribution as well. At least at the beginning, in the slowing down of the external field ramp. But in the long term, the field exclusion is due to the Meissner effect - the EM field actually becomes massive (the "photon" has a mass - it becomes a plasmon) so it cannot penetrate in the material. $\endgroup$
    – SuperCiocia
    Oct 11 '21 at 19:52
  • 1
    $\begingroup$ @SofiaAlfaro If I understand you correctly, then the answer is yes. If you hold an ideal conductor in your hand and bring a magnet toward it, the changing magnetic field will induce a current along the surface of the conductor which serves to exactly cancel the magnetic field in the interior; this is an effect of Lenz's law (or Faraday's law, if you prefer). The magnetic field inside a perfect conductor must be constant, so if it is initially zero then the induced surface currents will keep it that way. As SuperCiocia says though, this fails to explain the expulsion of the field at $T_c$. $\endgroup$
    – J. Murray
    Oct 12 '21 at 0:58
  • $\begingroup$ Thank yo for answering! I have a question for superciocia if it is okey! You wrote "The reason for this expulsion is due to the superconductor going through a phase transition in which the microscopic description of the material changes. It is well beyond high school level, but the electromagnetic field propagates in a different way when the material becomes superconducting, and it follows the London equations, which show an exponential damping of the field" Would you mind explaining this a little bit more, specially the part about "the phase transition". $\endgroup$ Oct 20 '21 at 17:48
  • 1
    $\begingroup$ All I meant is that a superconductor is not “just” an ideal conductor. There is actually a phase transition at a critical temperature, same as between ice and liquid water. So it’s a different phase. The microscopic description changes, and this new description results in the complete expulsion of the magnetic filed from the bulk. So even using a “perfect diamagnetism” analogy isn’t correct. $\endgroup$
    – SuperCiocia
    Oct 20 '21 at 19:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.